All Answers 358187 Answers

[-2,7] would be the notation

By the distance formula, we are trying to minimize \(\sqrt{x^2+y^2}=\sqrt{x^2+x^4-10x^2+25}\). In general minimization problems like this require calculus, but one elementary optimization method that sometimes works is completing the square. We have\(\sqrt{x^2+x^4-10x^2+25}=\sqrt{(x^2-9/2)^2+(25-81/4)}.\)

This expression is minimized when the square equals \(0\), i.e. when \(x=\pm 3/\sqrt{2}.\) For this value of \(x\), the distance is \(\sqrt{25-\frac{81}{4}}=\frac{\sqrt{19}}{2}.\)  Hence the desired answer is \(19+2=\boxed{21}\).

See the following image: 

We are looking for the area [ AEMF]

Let B  = (0, 0)   C  = (4, 0)   A  =  (2, 2√3)

Angle EMC  = 45° 

The segment MN  lies  on a line with a slope  = tan45°  = 1

So....the equation of this iine is   y = ( 1) (x - 2)  =  x - 2    (1) 

And the segment  AC   lies on a line with a slope  of tan (120°)  =  -√3 ( x - 4) = -√3 x + 4√3   (2)

So......the x value of the intersection of (1) and (2)     can be found as

x - 2  =  -√3 x + 4√3

x ( 1 + √3) = 4√3 + 2

x  = [ 4√3 + 2 ] / [ 1 + √3]

So   the y value  of  this intersection is

y =    [ 4√3 + 2 ] / [ 1 + √3]  - 2   =   [ 2√3 ] / [ 1 + √3]

So  E  =   ( [ 4√3 + 2 ] / [ 1 + √3] ,  [ 2√3 ] / [ 1 + √3] )

So...the area of triangle EMC  = (1/2) ( MC) ([ 2√3 ] / [ 1 + √3]) =  (1/2)(2)[ 2√3 ] / [ 1 + √3] =

[ 2√3 ] / [ 1 + √3] units^2     (3)     

And the area of  AMC  =  (1/2)(2√3)(2)  = 2√3  units^2   (4)

So   area  [ AEMF]  =   2 ( (4) - (3)  )   =    2  ( 2√3  -  [ 2√3 ] / [ 1 + √3]  )  =  2 (3√3 - 3)  units^2  ≈ 4.39 units^2

sorry...

okay melody i will try to do this next time

although i have forgoten what i had done

your wellcome

so is -2 to 7 what I put in? I am on AoPS this stuff is really confusing.

[24^50 - 15^50 ]  / 13   =  

[ 24^25 + 15^25 ] [ 24^25 - 15^25] / 13    = 

[ (8 * 3)^25 + (5 * 3)^25 ]  [ (8 * 3)^25 - (5 * 3)^25 )  / 13  =

[ 8^25 * 3^25 + 5^25 * 3^25 ]  [ 8^25 * 3^25 - 5^25 * 3^25 ]  / 13   = 

[ 3^25  ( 8^25 + 5^25) ]  [ 3^25 (  8^25 - 5^25)] / 13

[ 3^50 ] [ 8^25 + 5^25 ] [ 8^25 - 5^25 ]  / 13

Note that

(8^5) mod 13  = 8

And by a modular math property, since 5 is a factor of 25....then (8^25)mod 13  = 8

Also

(5^5) mod 13  = 5

And......using the same property......(5^25) mod 13  = (5^5) mod 13  =  5

So [ 8^25 + 5^25 ] /13   =  [ 8^25 mod 13 + 5^25 mod 13 ] /13  =

[ 8mod 13 + 5 mod 13 ]/13  =  [8 + 5] /13  = 13 /13  =  13 mod 13  =  0 = remainder

So

[ 3^50] * [ 8^25 + 5^25]  * [ 8^25 - 5^25]

             ____________

                     13

Is evenly divisible by 13...so

(24*50 - 15^50)   is  evenly divisible by 13.....so.....the remainder   =  0

Subtract 10 from both sides of the equation to get

x^2-5x-14 <=0

(x-7)(x+2) <= 0

x =7 and -2 for thequation to EQUAL 0       intervals  :   -inf to -2    -2 to 7   and 7 to +inf

                                                                             which do you think satisfy the equation?

JUst use the calculator on this website:

cos(3.9269908169872415)*1.6 = -1.1313708498992

y = x^2 + a

y = ax

Set the y's equal

x^2 + a  =  ax          rearrange as

x^2 - ax + a = 0 

For this to have real solutions.......the discriminant must  be ≥  0

So

a^2  -  4a ≥  0

a ( a - 4) ≥  0

Setting each factor to 0  and solving for a  produces the following solutions  a = 0   and a = 4

So  we have the following possible intervals that  will produce solutions

(-inf, 0 ]  or  ( 0 , 4 )   or  [ 4, inf )

If a  is in the first interval, then a (a - 4) ≥ 0  so this interval produces a solution

If  a is in the second interval, then a(a - 4)  < 0.....so  no solutions are found here

If a  is in the third interval, then a (a - 4) ≥ 0 .....so this interval produces a solution

So....the solution intervals are  

(-inf, 0 ]  U [ 4, inf )...as EP found  !!!

Let e(x) be an even function, and let o(x) be an odd function, such that

\(e(x) + o(x) = \frac{6}{x + 2} + x^2 + 2^x \)

for all real numbers x. Find o(1).

\(e(x)=x^2+2^x\ \{even\ function \}\\ o(x)=\frac{6}{x+2}\ \{odd\ function\}\\ \color{blue} o(1)=\frac{6}{1+2}=2\)

\(e(x) + o(x) =(x^2+2^x)+( \frac{6}{x + 2} )\)

  !

For each positive integer n,
the set of integers \(\{0,\ 1,\ 2,\ \ldots \,\ n-1 \}\) is known as the residue system modulo n.
Within the residue system modulo \(2^4\),
let A be the sum of all invertible integers modulo \(2^4\) and
let B be the sum all of non-invertible integers modulo \(2^4\).
What is A-B?

The integer s of the set is invertible, if \(gcd(2^4,s)=1\)

\(\begin{array}{|c|c|c|l|} \hline \text{set of integers} & gcd(2^4,s) & \text{not invertible} & \text{modulo inverse} \\ \hline 0 & 16 & \checkmark \\ \hline 1 & 1 && 1^{-1} \pmod{16} = 1 \quad |\quad 1\cdot 1 \equiv 1 \pmod{16} \\ \hline 2 & 2 & \checkmark \\ \hline 3 & 1 && 3^{-1} \pmod{16} = 11 \quad |\quad 3\cdot 11 \equiv 1 \pmod{16} \\ \hline 4 & 4 & \checkmark \\ \hline 5 & 1 && 5^{-1} \pmod{16} = 13 \quad |\quad 5\cdot 13 \equiv 1 \pmod{16} \\ \hline 6 & 2 & \checkmark \\ \hline 7 & 1 && 7^{-1} \pmod{16} = 7 \quad |\quad 7\cdot 7 \equiv 1 \pmod{16} \\ \hline 8 & 8 & \checkmark \\ \hline 9 & 1 && 9^{-1} \pmod{16} = 9 \quad |\quad 9\cdot 9 \equiv 1 \pmod{16} \\ \hline 10 & 2 & \checkmark \\ \hline 11 & 1 && 11^{-1} \pmod{16} = 3 \quad |\quad 11\cdot 3 \equiv 1 \pmod{16} \\ \hline 12 & 4 & \checkmark \\ \hline 13 & 1 && 13^{-1} \pmod{16} = 5 \quad |\quad 13\cdot 5 \equiv 1 \pmod{16} \\ \hline 14 & 2 & \checkmark \\ \hline 15 & 1 && 15^{-1} \pmod{16} = 15 \quad |\quad 15\cdot 15 \equiv 1 \pmod{16} \\ \hline \hline \end{array}\)

\( A = 1+3+5+7+9+11+13+15 = \mathbf{64} \\ B = 0+2+4+6+8+10+12+14 = \mathbf{56}\)

ok thanks Chris.

How about [4. inf) ?         and    (-inf, 0]  ?

Simplify the following:
1/(1 + sqrt(3) + sqrt(5)

Multiply numerator and denominator of 1/(1 + sqrt(3) + sqrt(5)) by 1 - sqrt(3) - sqrt(5):
(1 - sqrt(3) - sqrt(5))/((1 + sqrt(3) + sqrt(5)) (1 - sqrt(3) - sqrt(5)))

(1 + sqrt(3) + sqrt(5)) (1 - sqrt(3) - sqrt(5)) = 1 (1 - sqrt(3) - sqrt(5)) + sqrt(3) (1 - sqrt(3) - sqrt(5)) + sqrt(5) (1 - sqrt(3) - sqrt(5)) = (1 - sqrt(3) - sqrt(5)) + (sqrt(3) - 3 - sqrt(15)) + (sqrt(5) - sqrt(15) - 5) = -2 sqrt(15) - 7:
(1 - sqrt(3) - sqrt(5))/(-2 sqrt(15) - 7)

Factor -1 out of -7 - 2 sqrt(15) giving -(2 sqrt(15) + 7):
(1 - sqrt(3) - sqrt(5))/(-(2 sqrt(15) + 7))

Multiply numerator and denominator of (1 - sqrt(3) - sqrt(5))/(-(2 sqrt(15) + 7)) by -1:
-(1 - sqrt(3) - sqrt(5))/(2 sqrt(15) + 7)

-(1 - sqrt(3) - sqrt(5)) = -1 + sqrt(3) + sqrt(5):
(-1 + sqrt(3) + sqrt(5))/(2 sqrt(15) + 7)

Multiply numerator and denominator of (-1 + sqrt(3) + sqrt(5))/(2 sqrt(15) + 7) by 2 sqrt(15) - 7:
((-1 + sqrt(3) + sqrt(5)) (2 sqrt(15) - 7))/((2 sqrt(15) + 7) (2 sqrt(15) - 7))

(2 sqrt(15) + 7) (2 sqrt(15) - 7) = 7 (-7) + 7×2 sqrt(15) + 2 sqrt(15) (-7) + 2 sqrt(15)×2 sqrt(15) = -49 + 14 sqrt(15) - 14 sqrt(15) + 60 = 11:

((-1 + sqrt(3) + sqrt(5)) (2 sqrt(15) - 7))/11 = 1/11 (7 + 3 sqrt(3) - sqrt(5) - 2 sqrt(15)) =a + b + c + d =7 + 3 - 1 - 2 = 7

thanks travisio

\((4, \inf)\) 

Simplify the following:
sqrt(5) + 1/sqrt(5) + sqrt(7) + 1/sqrt(7)

Rationalize the denominator. 1/sqrt(5) = 1/sqrt(5)×(sqrt(5))/(sqrt(5)) = (sqrt(5))/5:
sqrt(5) + (sqrt(5))/5 + sqrt(7) + 1/sqrt(7)

Rationalize the denominator. 1/sqrt(7) = 1/sqrt(7)×(sqrt(7))/(sqrt(7)) = (sqrt(7))/7:
sqrt(5) + (sqrt(5))/5 + sqrt(7) + (sqrt(7))/7

Put each term in sqrt(5) + (sqrt(5))/5 + sqrt(7) + (sqrt(7))/7 over the common denominator 35: sqrt(5) + (sqrt(5))/5 + sqrt(7) + (sqrt(7))/7 = (35 sqrt(5))/35 + (7 sqrt(5))/35 + (35 sqrt(7))/35 + (5 sqrt(7))/35:
(35 sqrt(5))/35 + (7 sqrt(5))/35 + (35 sqrt(7))/35 + (5 sqrt(7))/35

(35 sqrt(5))/35 + (7 sqrt(5))/35 + (35 sqrt(7))/35 + (5 sqrt(7))/35 = (35 sqrt(5) + 7 sqrt(5) + 35 sqrt(7) + 5 sqrt(7))/35:
(35 sqrt(5) + 7 sqrt(5) + 35 sqrt(7) + 5 sqrt(7))/35

Add like terms. 35 sqrt(5) + 7 sqrt(5) + 35 sqrt(7) + 5 sqrt(7) = 42 sqrt(5) + 40 sqrt(7):

(42 sqrt(5) + 40 sqrt(7))/35  = a + b + c =42 + 40 + 35 = 117

thanks!

OK! Here is my best understanding of this:

A=1 + 3 + 5 + 7 + 9 + 11 + 13 + 15=64 sum of all  invertible integers mod 2^4.
B=2 + 4 + 6 + 8 + 10 + 12 + 14      =56 sum of all non-invertible integers mod 2^4.
A - B =64 - 56 = 8

Note: Somebody should check this. heureka maybe?

1.6cos(pi/0.8) = -1.1313708498992 

this may be rong i do not know how to do that!

thanks hectictar

x² + 5x  <  6

                            Subtract  6  from both sides of the inequality.

x² + 5x - 6  <  0

                            Let's find what values of  x  make  x² + 5x - 6  equal  0

x² + 5x - 6  =  0

                            Factor the left side. What two numbers add to  5  and multiply to  -6 ?   -1  and  +6

(x - 1)(x + 6) = 0

                            Set each factor equal to zero and solve for  x

1) 9/100

2) 5